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Solving
Spectroscopy
Problems:
Putting
it
All
Together
Once
you’ve
analyzed
the
mass
spectrometry,
infrared
spectrometry,
1H‐NMR,
and
13C‐
NMR
data,
there
is
no
one
way
to
put
them
together.
It’s
all
about
trial
and
error,
but
here
are
a
few
helpful
tips
and
tricks
to
help
you
get
on
the
right
track.
First,
here’s
an
overview
of
the
information
you
should
have
obtained
based
on
your
analyses
of
the
data
from
the
4
methods
of
spectrometry.
Mass
Spectrometry‐
molecular
formula,
DBE
Infrared
Spectrometry‐
present
functional
groups
1
H‐NMR‐
backbone
structure
(C‐H)
13
C‐NMR‐
hydrogens
attached
to
the
carbons,
confirms
functional
groups
found
in
IR
and
implications
determined
by
H‐NMR
Key
things
to
remember
as
you
proceed
with
the
puzzle:
1) Keep
track
of
the
number
atoms
in
your
molecule.
Make
sure
the
sum
of
the
atoms
in
your
1H‐NMR
implications
+
functional
groups
from
IR
spectrum
matches
the
molecular
formula
obtained
from
the
mass
spectrum.
2) Keep
track
of
the
DBE
count,
especially
when
doing
IR
analysis.
3) Check
that
your
implications
and
your
final
molecule
match
1H‐NMR
and
13C‐
NMR
chemical
shifts.
4) Check
that
your
final
molecule
matches
the
number
of
signals,
splitting
pattern,
and
number
of
hydrogen
atoms
as
the
1H‐NMR
and
13C‐NMR
data
The
information
from
the
mass
spectrum
is
pretty
straight
forward.
Use
it
to
keep
track
of
the
number
of
each
atom
as
you
proceed
with
the
IR
analysis
and
the
1H‐NMR
spectrum
(number
of
carbons
and
hydrogens).
Here’s
a
handy
table
that
will
help
you
out
with
determining
the
functional
group
associated
with
the
given
carbonyl
group:
1
13
Functional
Group
IR
H‐NMR
C‐NMR
Aldehyde
2
C‐H
peaks:
~2900
and
Aldehyde
proton
9.5‐
C=O
carbon
doublet
~2700
cm‐1
11
ppm
180‐220
ppm
C=O
stretch
1740‐1720
cm‐1
Ketone
C=O
stretch
1750‐1705
No
characteristic
1H
C=O
carbon
singlet
‐1
cm
chemical
shift
180‐220
ppm
1
Ester
C=O
stretch
1750‐1735
No
characteristic
H
C=O
carbon
singlet
‐1
cm
chemical
shift
160‐180
ppm
Carboxylic
acid
Broad
O‐H
stretch
3000‐ COOH
proton
10‐13
C=O
carbon
singlet
‐1
2500
cm
ppm
175‐185
ppm
C=O
stretch
1725‐1700
cm‐1
Amide
N‐H
stretch
3500‐3300
N‐H
proton
5‐9
ppm
cm‐1
(2
peaks
if
1°,
1
peak
if
2°,
none
if
3°)
C=O
stretch
1690‐1650
cm‐1
C=O
carbon
singlet
160‐180
ppm
Dr.
Hardinger’s
Chemistry
14C
Thinkbook
,
Ninth
Edition
First,
list
out
all
implications,
functional
groups,
and
remaining
atoms
that
you
have
found
from
the
IR,
1H‐NMR,
and
13C‐NMR
data.
Consolidate
as
many
1H‐NMR
implications
from
different
signals
as
possible
(will
be
discussed
shortly).
Once
that’s
settled,
a
good
starting
place
is
to
draw
out
the
functional
groups
found
in
the
IR
spectrum
such
as:
Then,
use
trial
and
error
to
attach
the
remaining
pieces
(1H‐NMR
and
IR
functional
groups).
For
1H‐NMR,
here
are
some
hints
and
patterns
that
occur
frequently
(but
not
always):
But
first,
here’s
a
table
of
the
more
common
implications
and
the
ones
that
are
most
likely
to
make
up
part
of
the
final
molecule
(i.e.
one
that
show
up
in
the
Thinkbook
and
practice
exams).
It
does
not
show
all
possible
implications.
………
these
are
often
overlooked,
but
keep
these
implications
in
mind
when
there
are
oxygens
(alcohols
more
specifically)
and
nitrogens
in
the
molecular
formula
and
when
alcohols,
amines,
carbonyls,
carboxylic
acids,
and
aldehydes
are
present
it
the
IR
spectrum
………
these
atoms
represent
the
carbons
with
the
correct
number
of
hydrogens
indicated
by
the
integration
Shift
Signal
Integration
Number
of
Implications
Hydrogens
singlet
1
1
CH,
OH,
NH,
C=O,
CHO,
COOH
2
2
CH2,
CH
x
2
3
3
CH3
CH
x
3
4
4
CH2
x
2,
CH
x
4
6
6
CH3
x
2,
CH2
x
3,
CH
x
6
9
9
CH3
x
3,
CH
x
9
Doublet
1
1
CHCH
2
2
CHCH2
3
3
CHCH3
4
4
CHCH2,
6
6
CHCH3
x
2
9
9
CHCH3
x
3
Triplet
1
1
CH2CH,
CHCHCH
2
2
CH2CH2,,
CHCH2CH
3
3
CH2CH3,
CHCH3CH
4
4
CH2CH2
x
2,
CHCH2CH
x
2
6
6
CH2CH3
x
2,
CHCH3CH
x
2
9
9
CH2CH3
x
3,
CHCH3CH
x
3
Quartet
1
1
CH3CH,
CH2CHCH
2
2
CH3CH2,,
CH2CH2CH
3
3
CH3CH3,
CH2CH3CH
4
4
CH3CH2
x
2,
CH2CH2CH
x
2
6
6
CH3CH3
x
2,
CH2CH3CH
x
2
9
9
CH3CH3
x
3,
CH2CH3CH
x
3
Pentet
1
1
CH3CHCH,
CH2CHCH2
2
2
CH3CH2CH,,
CH2CH2CH2
3
3
CH3CH3CH,
CH2CH3CH2
4
4
CH3CH2CH
x
2,
CH2CH2CH2
x
2
6
6
CH3CH3CH
x
2,
CH2CH3CH2
x
2
9
9
CH3CH3CH
x
3,
CH2CH3CH2
x
3
Sextet
1
1
CH3CHCH2,
CH2CHCH2
2
2
CH3CH2CH2,
CH2CH2CH3
3
4
6
9
Septet
1
2
3
4
6
9
Multiplet
5
2
4
Doublets
3
4
6
9
1
2
3
4
6
9
5
4
CH3CH3CH2,
CH2CH3CH3
CH3CH2CH2
x
2
CH3CH3CH2
x
2
CH3CH3CH2
x
3
CH3CHCH3
CH3CH2CH3
CH3CH3CH3
CH3CH2CH3
x
2
CH3CH3CH3
x
2
CH3CH3CH3
x
3
C6H5
monosubstituted
benzene
ring
C6H4
disubstituted
benzene
ring
Multiplets
and
“doublets
of
doublets”
are
usually
benzene
rings
for
our
purposes.
Implications
of
the
form
CH3
x
2
or
3
are
often
present
in
molecules
with
2
or
3
CH3
groups
attached
to
the
same
atom
or
show
symmetry
in
other
ways
across
the
molecule.
i.e.
CH3
x
2
:
CH3
x
3
:
Often
times
you
can
couple
together
implications
for
several
signals
if
they
coincide
with
each
other.
i.e.
a
triplet
CH2CH3
and
a
quartet
CH3CH2
can
be
put
together
into
one
single
piece:
CH2CH3
When
you
need
to
separate
implications
for
different
signals
so
that
they
don’t
couple,
try
inserting
these
atoms/molecules
in
between
them:
O
C=O
N
i.e.
CH3CH2CH2CH2CH2
Gives:
2
triplets,
2
pentets,
1
sextet
whereas
CH3CH2OCH2CH2CH2
Gives:
3
triplets,
1
quartet,
and
1
pentet
Now
let’s
put
it
all
together
with
some
practice
problems!
13
C‐NMR,
2D‐NMR,
and
MRI
OWL
#4.
Suggest
the
structure
for
a
molecule
of
formula
C10H14O
with
the
following
NMR
data.
Formula:
C10H14O
DBE
=
10
–
(14/2)
+
(1/0)
+
1
=
4
(probably
benzene
ring)
1
H‐NMR:
7.12
ppm
(doublet
of
doublets;
integral=
2),
6.82
ppm
(doublet
of
doublets;
integral
=
2),
3.74
ppm
(singlet;
integral
=
3),
2.84
ppm
(septet;
integral
=
1),
and
1.21
ppm
(doublet;
integral
=
6).
13
C‐NMR:
157.9
ppm
(singlet),
141.1
ppm
(singlet),
127.3
ppm
(doublet),
113.9
ppm
(doublet),
55.2
ppm
(quartet),
33.4
ppm
(doublet),
and
24.2
ppm
(quartet)
After
completion
of
the
IR
and
NMR
analyses,
you
should
have
the
following
pieces:
1
H‐NMR:
C6H4
disubstituted
benzene
ring
CH3
singlet
CH(CH3)2
CH(CH3)2
Here
I’ve
matched
the
proton
CH3CH
x
2
13
and
carbon
NMR
implications
C‐NMR:
by
highlighting
them
with
the
157.9
ppm
singlet
–benzene
ring
C
same
color.
You
can
see
that
141.1
ppm
singlet
–
benzene
ring
C
13
C‐NMR
confirms
the
1H‐
127.3
ppm,
113.9
ppm
–2
CH
on
benzene
ring
NMR
implications
55.2
ppm
quartet
–
CH3
(probably
directly
attached
to
benzene
ring)
33.4
ppm
doublet
–
CH
(probably
directly
attached
to
benzene
ring)
24.2
ppm
quartet
–
CH3
KEEP
TRACK
of
the
number
of
atoms
in
your
molecule!
C6H4+
CH(CH3)2
+
CH3
=
C10H14
What’s
left:
O
1) start
by
drawing
the
benzene
ring:
2) Since
the
benzene
ring
is
isubstiuted
there
are
2
carbons
on
it
that
can
connect
to
the
2
remaining
pieces
of
the
puzzle.
These
2
pieces
can
be
arranged
in
3
possible
ways:
Since
there
are
4
13C‐NMR
signals
attached
to
the
benzene
ring.
Only
the
para
isomer
gives
this
arrangement
fits
our
structure.
3) Since
we
still
have
an
oxygen
left,
it
can
attach
to
either
the
CH
or
the
CH3
singlet.
Look
at
the
1H‐NMR
chemical
shift
and
you
will
see
that
the
CH3
singlet
has
the
higher
chemical
shift,
so
it
is
probably
attached
to
the
oxygen.
So,
attach
the
CH(CH3)2
on
one
position
on
the
benzene
ring
and
the
CH3O
you
just
made
on
the
other
position.
Make
sure
to
check
you
have
all
the
required
atoms
and
check
for
the
correct
number
of
NMR
signals
and
the
correct
chemical
shifts.
And
there
you
have
it!
Let’s
try
a
harder
one:
13
C‐NMR,
2D‐NMR,
and
MRI
Practice
Problem
#16:
Deduce
the
structure
that
corresponds
to
the
spectral
data.
Mass
Spectrum:
m/z
242
(M:
100%),
m/z
243
(17.87%),
and
m/z
132
(0.50%)
Solve
for
formula:
C16H18O2
DBE
=
8
IR:
alcohol
O‐H,
aryl/vinyl
sp2
C‐H,
alkyl
sp3
C‐H,
benzene
ring
1
H‐NMR:
7.24‐7.14
ppm
(multiplet;
integral
=
5),
3.62
ppm
(triplet;
integral
=
1),
2.50
ppm
(singlet;
integral
=
1)
1.68
ppm
(pentet;
integral
=
1),
and
1.51
ppm
(triplet;
integral
=
1)
13
C‐NMR:
142.3
ppm
(singlet),
128.4
ppm
(doublet),
128.3
ppm
(doublet),
125.7
ppm
(doublet),
78.5
ppm
(singlet),
62.0
ppm
(triplet),
39.5
ppm
(triplet),
and
27.2
ppm
(triplet)
Pieces:
2
x
C6H5
monosubstituted
benzene
ring
CH3CH2CH
or
CH2CH2CH2
CH2CH2
Put
these
3
2
x
OH
CH2CH2
together
leftovers:
O,
C
CH2CH2CH2
1) Since
the
2
benzene
rings
are
equivalent
(same
signal),
draw
out
the
benzene
rings
next
to
each
other
so
they
can
somehow
be
equivalent.
2) It
seems
you
can’t
attach
them
with
the
remaining
pieces
in
between
them,
linking
them
together
as
the
CH2CH2CH2
,
C,
and
O
cannot
form
a
symmetric
chain
on
its
own.
The
leftover
C
must
attach
the
two
rings.
3) The
leftover
C
has
2
benzene
rings
attached,
so
it
has
2
more
bonds
to
form.
One
can
be
the
CH2CH2CH2
chain.
(this
one
fits
better
with
the
other
implications‐
CH2CH2)
4) The
other
must
be
an
OH
as
that
is
all
we
have
left.
5) The
other
OH
must
cap
off
the
end
of
the
CH2CH2CH2
chain
to
complete
the
molecule.
And
since
“The
two
ends
of
this
CH2CH2CH2
piece
are
not
equivalent”
the
molecule
must
be
constructed
this
way.
(Dr.
Hardinger’s
Chemistry
14C
Thinkbook
,
Ninth
Edition)
Remember
again
to
check
your
work‐
molecular
formula,
IR
functional
groups,
NMR
signals,
integrals,
and
chemical
shifts.
References:
Dr.
Hardinger’s
Chemistry
14C
Thinkbook
,
Ninth
Edition
(pgs.
217,
221‐222,
229,
233‐235)
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